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denpristay [2]
2 years ago
8

Identify any incorrect formulas. Explain your answer.

Chemistry
1 answer:
stepladder [879]2 years ago
5 0
A is incorrect because Mg has a 2+ charge, while SO4 has a 2- charge, if there is 2 Mg and 3 SO4 then the formula will not add to 0 meaning neutral
You might be interested in
Calculate the mass of Octane needed to release 6.20 mol Co2
n200080 [17]
The combustion reaction of octane is as follow,

                           C₈H₁₈  +  25/2 O₂     →     8 CO₂  +  9 H₂O

According to balance equation,

8 moles of CO₂ are released when  =  114.23 g (1 mole) Octane is reacted

So,

      6.20 moles of CO₂ will release when  =  X g of Octane is reacted

Solving for X,
                                     X  =  (114.23 g × 6.20 mol) ÷ 8 mol

                                     X  =  88.52 g of Octane
Result:
           88.52 g of Octane is needed to release 6.20 mol CO₂.
8 0
2 years ago
Help please. Which mass of urea, CO(NH2)2, contains the same mass of nitrogen as 101.1 g of potassium nitrate?
Masja [62]
Did u mean mass of potassium nitrate or the mass of nitrogen in potassium nitrate

6 0
3 years ago
A calorimeter contains 22.0 mL of water at 14.0 ∘C . When 2.50 g of X (a substance with a molar mass of 82.0 g/mol ) is added, i
Alik [6]

Answer:

The enthalpy change in the the reaction is -47.014 kJ/mol.

Explanation:

X(s)+H_2O(l)\rightarrow X(aq)

Volume of water in calorimeter = 22.0 mL

Density of water = 1.00 g/mL

Mass of the water in calorimeter = m

m=1.00 g/mL\times 22.0 mL=22 g

Mass of substance X = 2.50 g

Mass of the solution = M = 2.50 g + 22 g = 24.50 g

Heat released during the reaction be Q

Change in temperature =ΔT = 28.0°C - 14.0°C = 14.0°C

Specific heat of the solution is equal to that of water :

c = 4.18J/(g°C)

Q=Mc\times \Delta T

Q=24.50 g\times 4.18 J/g ^oC\times 14.0^oC=1,433.74 J=1.433 kJ

Heat released during the reaction is equal to the heat absorbed by the water or solution.

Heat released during the reaction =-1.433 kJ

Moles of substance X= \frac{2.50 g}{82.0 g/mol}=0.03048 mol

The enthalpy change, ΔH, for this reaction per mole of X:

\Delta H=\frac{-1.433 kJ}{0.03048 mol}=-47.014 kJ/mol

5 0
2 years ago
PLEASE HELP ASAP !!
Ivenika [448]

Answer:

  • 2SO₂ +  O₂ + 2H₂O  -----------> 2H₂SO₄
  • Theoretical yield of H₂SO₄ = 213 g
  • percent yield of H₂SO₄ = 94 %  

Explanation:

Data Given:

volume of SO₂ = 48.6 L

mass of H₂SO₄ = 200 g

balance equation = ?

theoretical yield = ?

percent yield = ?

Solution:

Part 1:

first we have to write a balance equation for the reaction

SO₂ gas react with water (H₂O) and excess oxygen

The balanced equation is as under

                       2SO₂ +  O₂ + 2H₂O  -----------> 2H₂SO₄

Part 2:

Now we have to find theoretical yield

First look at the balance reaction

                        2SO₂ +  O₂ + 2H₂O  -----------> 2H₂SO₄

                        2 mol                                         2 mol

2 moles of SO₂ give gives 2 moles of H₂SO₄

Now calculate volume of 2 moles of SO₂ and mass of 2 moles of H₂SO₄

volume of 2 moles of SO₂

Formula used

                 volume of gas = no. of moles x molar volume . . . . . . (1)

molar volume of SO₂= 22.4 L/mol

Put values in above formula (1)

                 volume of gas = 2 mol x 22.4 L/mol

                 volume of gas = 44.8 L

volume of 2 mole of SO₂ = 44.8 L

Now,

Find mass of 2 mole H₂SO₄

Formula Used

            mass in grams = no. of moles x molar mass . . . . . . . (2)

molar mass of H₂SO₄ = 2 (1) + 32 + 4(16)

molar mass of H₂SO₄ = 98 g/mol

put values in equation 2

        mass in grams = 2 mol x 98 g/mol

        mass in grams = 196 g

mass of 2 mole of H₂SO₄ = 196 g

** So,

Now we come to know that

44.8 L of SO₂ gives 196 g of H₂SO₄ then how many grams of the H₂SO₄ will be produced by 48.6 L of SO₂

Apply unity Formula

               44.8 L of SO₂ ≅ 196 g of H₂SO₄

               48.6 L of SO₂ ≅ X g of H₂SO₄

Do cross multiplication

                g of H₂SO₄  = 196 g x 48.6 L / 44.8 L

                g of H₂SO₄  =  213 g

So that is why the theoretical yield of H₂SO₄ is 213 g

Theoretical yield of H₂SO₄ = 213 g

Part 3

Calculate Percent Yield:

Formula used for this purpose:

             percent yield = actual yield /theoretical yield x 100 %

Put value in the above formula

           percent yield = 200 g/ 213 g x 100 %

          percent yield = 94 %    

So percent yield of H₂SO₄ = 94 %    

8 0
2 years ago
B. If the sand you ran across has a specific-heat capacity of 835 J/(kgºc),
Murrr4er [49]

Answer: 16700 Joules

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

Q=m\times c\times \Delta T

Q = Heat absorbed = ?

m = mass of sand = 2 kg

c = heat capacity = 835J/kg^0C

Initial temperature  = T_i = 40^0C

Final temperature= T_f  = 50^0C

Change in temperature ,\Delta T=T_f-T_i=(50-40)^0C=10^0C

Putting in the values, we get:

Q=2kg\times 835J/kg^0C\times 10^0C

Q=16700J

16700 J of energy must be added to a 2-kilogram pile of it to increase its temperature from 40°C to 50°C

4 0
2 years ago
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